be an acute triangle with 
and let 
be its orthocenter. Let 
be a point on the perpendicular bisector of 
such that 
and 
are on different sides of 
,
a point on the perpendicular bisector of 
such that 
and 
a point on the perpendicular bisector of 
such that 
and 
lie on the opposite side of 
w.r.t 
.
and 
be a polynomial with integer coefficients such that 
,
be an integer. Define 
and 
for all integers 
.
such that 
.
such that 
coprime, 
and 
.
be the roots of the polynomial 
.
if within that number exists a digit 
that appears exactly 
times in that number, hence find the number of 
be the real roots of![\[x^3 + 7x^2 -6x - 13 = 0\]](http://latex.artofproblemsolving.com/6/6/3/6639be4c9e92b5ad4466359681db0d2a461ec8dd.png)
Find 
,
,
be the roots of 
Find the numerical value of the expression:![\[\frac{\alpha + 2}{\alpha - 2} + \frac{\beta + 2}{\beta - 2} + \frac{\gamma + 2}{\gamma - 2}\]](http://latex.artofproblemsolving.com/b/5/9/b59f4515409422333e10857ea775a36e58490b5d.png)
and 
such that 
and 
.
and 
such that:
,
is in 
,
and 
;
such that 
is not divisible by 
.
Y by nbasrl, cubres, farhad.fritl
Determine the sets 
of positive integers satisfying the following two conditions:
of positive integers satisfying the following two conditions:
- For any positive integers
, if 
is in 
, then so are 
and 
; and - The set contains an integer
such that 
is not divisible by 
.
.
,
with 
.
,
and 
both in 
,
modulo 
and 
,
.
.
,
.
.
for some 
.
.
,
.
for any 
.
.
then 
where 
is "generally smaller" than 
and 
is "generally larger" than 
,
impliying 
instead of 
and 
is divisible by 
except for 
and 
(as verified by a computer search) but I don't think a clean general argument can be made.
is crucial here and here's why a general argument for that case couldn't exist. Going back to the second point above, it can be proven that any number 
can be written as a sum 
except for at most 
numbers. These are: 
.
mod 
numbers on that list and it's not known if the 19th exists. If the Generalized Riemann Hypothesis is true, then there is no 19th number. See 
